Clapton coil question

[astronomicals]

I made a dual clapton coil that is 26 wrapped in 30. The resistance is the same is if I had not wrapped it with 30. Why is that? Whats happening?

its 5/6 wrap, 1/8" and it comes in at ~.5 ohms (with dual coils).

Is the 30 just acting as an insulator and being heated by the 26? Or vice versa?

 

[SLIPPY_EEL]

the flow of electric takes the path of least resistance, your thinner outer wire only heats up.

if you had split the two wires along the legs that enter the post holes then the current will pass through both wires but i think the resistance would then be something between your 0.5ohms and whatever your thinner wire is

 

[astronomicals]

Ok. That makes sense. It wasnt split. I guess the 30 just coincidentally was the same resistance as the 26 itself would have been.

Do you recommend a rebuild with the ends split?

 

[SLIPPY_EEL]

Do you recommend a rebuild with the ends split?

nope.. the only thing close that ive ever done is just a parallel build, i quite like having a parallel build where one strand is kanthal and the other nichrome or/and even dif gauge's, this makes a coil that as you fire it up you see every other wrap lighting slightly before the other, its called a staged coil>>https://www.youtube.com/watch?v=CzxaTKO_Clo

and a step up from this is doing a clapton over the top of the two strands making a fused clapton>> https://www.youtube.com/watch?v=X4MFwX0114k

another wire that changes in ohms as respect to wether you have twisted the legs or not before installing in the post holes is a simple 3 strand braid or any braid